A baseball is thrown downward from a 50-ft tower with an initial speed of 18 ft/s. Determine the speed at which it hits the ground and the time of travel.
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Solution:
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We will use the following formula to solve this question to figure out the speed at which it hits the ground.
(Where v_2 is final velocity, v_1 is initial velocity, a_c is constant acceleration, s_2 is initial displacement, and s_1 is final displacement)
From the question, we know the following values:
a_c=32.2 \,\text{ft/s}^2
s_2=50 ft
s_1=0 ft
Let us substitute these values into our equation:
v_2^2=18^2+2(32.2)(50-0)
v_2^2=3544
(Take the square root of both sides)
v_2=59.5 ft/s
To figure out the time of travel, we will use the following equation.
(Where v_2 is final velocity, v_1 is initial velocity, a is constant acceleration and t is time)
Again, substituting the values we know gives us:
(Solve for t)
t=1.29 s
Final Answers:
t=1.29 s
where you get the acceleration?
Hi!
Acceleration of gravity is 9.81 m/s^2, or 32.17 ft/s^2. This is a value given in most textbooks (and one that should be memorized because it will come up a lot). Hope that helps, and thanks for visiting our site 🙂
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32.2ft/s how do you get this????
This is given in the book, but also a value you must memorize. This is the acceleration due to gravity, and in SI terms, it’s 9.81 m/s^2. 🙂
How did you solve the time?
The steps are shown, please check through the solution.
how you get the acceleration =32.2
This is acceleration due to gravity, it’s in your book and any other physics book. In SI terms, its 9.81 m/s^2.